The exterior angles B and C in ∆ABC are bisected to meet at a point P.
Prove that angle BPC = 90°-A/2. Is ∆BPC a cyclic quadrilateral?
Answers
Answer:Given−
ΔABChasitsexternalangles∠CBR&∠BCQbisected
bythelinesBP&CPrespectively.
BP&CPintersectatP.
Tofindout−
(i)is∠BPC=90
o
−
2
A
?
(ii)isABPCacyclicquadrilateral?
Solution−
∠PBC=∠PBR⟹∠PBC+∠PBR=2∠PBC.
So∠ABC+2∠PBC=180
o
(linearpair).
i.e∠PBC=90
o
−
2
∠ABC
........(i).
Similarly∠PCQ=∠PCB⟹∠PCB+∠PCQ=2∠PCB.
So∠ACB+2∠PCB=180
o
(linearpair).
i.e∠PCB=90
o
−
2
∠ACB
........(ii).
Adding(i)&(ii)wehave
∠PBC+∠PCB=90
o
−
2
∠ABC
+90
o
−
2
∠ACB
=180
o
−
2
1
(∠ABC+∠ACB)
⟹180
o
−∠BPC=180
o
−
2
1
{180
o
−∠BAC}
⟹∠BPC=90
o
−
2
BAC
.......(ans−i).
Againthesumoftheoppositeanglesofthe
quadrilateralABPC
=∠BAC+∠BPC=∠BAC+90
o
−
2
BAC
=180
o
.
∴ThequadrilateralABPCisnotacyclic
quadrilateral.........(ans−ii).
(∵thesumoftheoppositeanglesofacyclic
quadrilateral=180
o
).
Ans−OptionB.
Step-by-step explanation:
Answer:
The exterior angles B and C in ABC are bisected to meet at a point P. Prove that BPC = 900
. Is ABPC a
cyclic quadrilateral ?
Step-by-step explanation:
The exterior angles B and C in ABC are bisected to meet at a point P. Prove that BPC = 900
. Is ABPC a
cyclic quadrilateral ?